Solve 4^x-(y+3)*2^x+3y-4=0

$${ 4 }^{ x } -(y+3) \times { 2 }^{ x } +3y-4=0$$

Solve for y

$$y=-\frac{4^{x}-3\times 2^{x}-4}{3-2^{x}}$$

$$x\neq \log_{2}\left(3\right)$$

Steps for Solving Linear Equation

Use the distributive property to multiply y+3 by 2^{x}.

$$4^{x}-\left(y\times 2^{x}+3\times 2^{x}\right)+3y-4=0$$

To find the opposite of y\times 2^{x}+3\times 2^{x}, find the opposite of each term.

$$4^{x}-y\times 2^{x}-3\times 2^{x}+3y-4=0$$

Subtract 4^{x} from both sides. Anything subtracted from zero gives its negation.

$$-y\times 2^{x}-3\times 2^{x}+3y-4=-4^{x}$$

Add 3\times 2^{x} to both sides.

$$-y\times 2^{x}+3y-4=-4^{x}+3\times 2^{x}$$

Add 4 to both sides.

$$-y\times 2^{x}+3y=-4^{x}+3\times 2^{x}+4$$

Combine all terms containing y.

$$\left(-2^{x}+3\right)y=-4^{x}+3\times 2^{x}+4$$

The equation is in standard form.

$$\left(3-2^{x}\right)y=3\times 2^{x}-4^{x}+4$$

Divide both sides by -2^{x}+3.

$$\frac{\left(3-2^{x}\right)y}{3-2^{x}}=\frac{3\times 2^{x}-4^{x}+4}{3-2^{x}}$$

Dividing by -2^{x}+3 undoes the multiplication by -2^{x}+3.

$$y=\frac{3\times 2^{x}-4^{x}+4}{3-2^{x}}$$

Solve for x

$$\left\{\begin{matrix}\\x=\log_{2}\left(\frac{\sqrt{y^{2}-6y+25}+y+3}{2}\right)\text{, }&\text{unconditionally}\\x=\log_{2}\left(\frac{-\sqrt{y^{2}-6y+25}+y+3}{2}\right)\text{, }&y>\frac{4}{3}\end{matrix}\right.$$